题目内容

(本题满分9分)如图所示,△ABC内接于⊙O,AB是⊙O的直径,点D在⊙O

上,过点C的切线交AD的延长线于点E,且AE⊥CE,连接CD.

(1)求证:DC=BC;  

(2)若AB=5,AC=4,求tan∠DCE的值.

 

(1)证明:连接OC················································································· 1分

∵OA=OC

∴∠OAC=∠OCA

∵CE是⊙O的切线

∴∠OCE=90° ·············································· 2分

∵AE⊥CE

∴∠AEC=∠OCE=90°

∴OC∥AE  ·················································· 3分 

∴∠OCA=∠CAD   ∴∠CAD=∠BAC

∴DC=BC  ··························································································· 4分

(2)∵AB是⊙O的直径  ∴∠ACB=90°

·························································· 5分

∵∠CAE=∠BAC  ∠AEC=∠ACB=90°

∴△ACE∽△ABC······················································································ 6分

    ∴   ······················································ 7分

 

∵DC=BC=3

····················································· 8分

 

-----------9分         (其它解法参考得分)

 

解析:略

 

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